Talk:Surgery obstruction map I (Ex)
Line 7: | Line 7: | ||
\mathrm{sign}(M)-\mathrm{sign}(X) & = \langle L(TM), [M] \rangle - \langle L(TX),[X]\rangle \\ | \mathrm{sign}(M)-\mathrm{sign}(X) & = \langle L(TM), [M] \rangle - \langle L(TX),[X]\rangle \\ | ||
& = \langle L(\nu_M)^{-1}, [M] \rangle - \langle L(TX),[X]\rangle \\ | & = \langle L(\nu_M)^{-1}, [M] \rangle - \langle L(TX),[X]\rangle \\ | ||
− | & = \langle L(\overline{f}^*(\eta))^{-1}, f_*([X]) \rangle \langle L(TX),[X]\rangle \\ | + | & = \langle L(\overline{f}^*(\eta))^{-1}, f_*([X]) \rangle - \langle L(TX),[X]\rangle \\ |
− | & = \langle f^*L(\eta)^{-1}, f_*([X]) \rangle \langle L(TX),[X]\rangle \\ | + | & = \langle f^*L(\eta)^{-1}, f_*([X]) \rangle - \langle L(TX),[X]\rangle \\ |
& = \langle L(\eta)^{-1},[X]\rangle - \langle L(TX),[X]\rangle | & = \langle L(\eta)^{-1},[X]\rangle - \langle L(TX),[X]\rangle | ||
\end{array}$$ | \end{array}$$ | ||
Line 14: | Line 14: | ||
Now $(\xi\oplus\xi,\phi * \phi)$ is the sum of $(\xi,\phi)$ and $(\xi,\phi)$ in $\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum. | Now $(\xi\oplus\xi,\phi * \phi)$ is the sum of $(\xi,\phi)$ and $(\xi,\phi)$ in $\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum. | ||
Moreover | Moreover | ||
− | $$\theta(-(\xi, | + | $$\theta(-(\xi,\phi))+\theta(-(\xi,\phi)) - \theta(-(\xi\oplus\xi,\phi * \phi)) |
= 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -\langle L(TX),[X]\rangle $$ | = 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -\langle L(TX),[X]\rangle $$ | ||
If this is non-zero, then the surgery obstruction is not a group homomorphism with respect to the Whitney sum. | If this is non-zero, then the surgery obstruction is not a group homomorphism with respect to the Whitney sum. |
Latest revision as of 18:37, 31 May 2012
If is a manifold, then the normal map gives the base point of . An element of is given by a bundle together with a fiber homotopy trivialization . Under the isomorphism , the pair corresponds to a normal map covered by . Assume that the dimension is and that is simply connected. Then the surgery obstruction of a normal map covered by equals
by the Hirzebruch signature theorem and several properties of the -genus. In particular the surgery obstruction depends only on the bundle over . Now is the sum of and in with respect to the Whitney sum. Moreover
If this is non-zero, then the surgery obstruction is not a group homomorphism with respect to the Whitney sum.
As an example take :
There are fiber homotopically trivial bundles on corresponding to classes in which restrict to any given class in , as follows from the Puppe sequence with . From another exercise we know that on we have such vector bundles with first Pontryagin class times the generator of . This means that on we have a vector bundle with whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence . We compute
where the constant can be computed from the L-genus to be .
So the surgery obstruction is not a group homomorphism with respect to the Whitney sum.